Arectangular hole is to be cut in a wall for a vent. if the perimeter of the hole is 112 in. and the length of the diagonal is a minimum, what are the dimensions of the hole? let the length and width of the hole be x and y, respectively, and let the length of the diagonal be d. find an equation that describes the quantity to be minimized. upper d squared equals x squared plus (56 minus x )squared (type an equation.) the length of the diagonal is a minimum when the hole is 28 in. by 28 in.

D = √ (2x² -112x + 3136

Step-by-step explanation:

From problem statement we have:

Perimeter of the hole      P = 112 in

length of the hole    x

wide of the hole       y

Perimeter is by definition     P  =  2*x + 2*y  so  y = [( P - 2* x)] ÷ 2

y = ( P - 2*x ) ÷ 2   and P = 112 in    

y = ( 112 - 2*x) ÷ 2   ⇒ y =  56 - x      (1)

Triangle ABC (see attached file)  is straight in A.  Then   diagonal D is:

D² = x²  + y²        from equation   (1)     D²  = x² + ( 56 - x )²      

Solving:      D² = x² +  3136 + x² - 112*x ⇒  D² = 2*x² - 112*x + 3136

Finally    D = √ (2x² -112x + 3136